# Finding the equation of a cubic when given $4$ points

I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$.

I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as.

If anyone could provide their working out for this problem it would be extremely enlightening.

the general cubic equation is $$y=ax^3+bx^2+cx+d.$$Plug in the coordinates of the points for x and y, and you end up with a system of four equations in four variables, namely $a, b, c$ and $d$. Hope that helps!

• I have done as you said and I have ended up with an equation for a=, b=, c=, and d=0(given). Where should I go next? The there are two variables on the RHS of these equations and I don't know how to proceed further. – Haotian Huang Feb 18 '18 at 7:13
• Use row operations to isolate the variables. For example, get an expression for a by isolating it and substitute it into another, until you get equations in one variable.(Or, if you know how, set up a matrix and row-reduce it.) – Ranjeev Grewal Feb 18 '18 at 7:17
• I've finally done it! Thank you very much Ranjeev – Haotian Huang Feb 18 '18 at 7:55

Given four points $(x_i,y_i)$ consider the functions $$f_1(x)=\frac {(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$ so that $f_1(x_1)=1$ and $f_1(x_i)=0, i\neq 1$, and similarly $f_2, f_3, f_4$. Note that the $f_i$ are cubic in $x$.

Then $p(x)=y_1f_1(x)+y_2f_2(x)+y_3f_3(x)+y_4f_4(x)$ is at most a cubic polynomial and passes through the four given points.

• See Lagrange Interpolation/Polynomials en.wikipedia.org/wiki/Lagrange_polynomial Note also that this is not the only place where a system with $f_i(x_j)=1(i=j) =0 (i\neq j)$ is useful. See for example (versions of) the Chinese Remainder Theorem. – Mark Bennet Feb 18 '18 at 15:01

Guide:

Let the equation be $y=ax^3+bx^2+cx+d$, since it passes through $(1,3)$, we have

$$a(1)^3+b(1)^2+c(1)+d=3$$

Do the same thing for the other $3$ points.

Hence you will obtain $4$ linear equation with $4$ variables.

You can then solve it using elementary row operations to recover $a,b,c,d$.

• Unfortunately, I have not learn elementary row operations yet. However I have obtained 4 linear equations with 4 variables. Is there a way to find the cubic equation without elementary row operations? – Haotian Huang Feb 18 '18 at 7:16
• what about substitution? – Siong Thye Goh Feb 18 '18 at 7:16

To determine a conic we need to solve 5 equations with 5 given points.

Likewise here we are given 4 points and 4 simultaneous equations, not involving any $xy$ term.. so solve it by Cramer's determinants.

An equation of the cubic that passes through the four points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and $(x_4,y_4)$ is $$\begin{vmatrix} x^3 & x^2 & x & y & 1 \\ x_1^3 & x_1^2 & x_1 & y_1 & 1 \\ x_2^3 & x_2^2 & x_2 & y_2 & 1 \\ x_3^3 & x_3^2 & x_3 & y_3 & 1 \\ x_4^3 & x_4^2 & x_4 & y_4 & 1 \end{vmatrix} = 0.$$ Plug in the coordinates of your points and simplify.

Hint:

It is proper to use the Lagrange function as following$$f(x)=\sum_{cyc}\dfrac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$